Problem: If
\[\tan x = \frac{2ab}{a^2 - b^2},\]where $a > b > 0$ and $0^\circ < x < 90^\circ,$ then find $\sin x$ in terms of $a$ and $b.$
Answer: We can construct a right triangle with legs $a^2 - b^2$ and $2ab.$  Then by Pythagoras, the hypotenuse is
\[\sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4 + 2a^2 b^2 + b^4} = a^2 + b^2.\][asy]
unitsize(1.5 cm);

pair A, B, C;

A = (2,1.8);
B = (0,0);
C = (2,0);

draw(A--B--C--cycle);
draw(rightanglemark(A,C,B,5));

label("$x$", B + (0.5,0.2));
label("$a^2 - b^2$", (B + C)/2, S);
label("$2ab$", (A + C)/2, E);
label("$a^2 + b^2$", (A + B)/2, NW);
[/asy]

Hence,
\[\sin x = \boxed{\frac{2ab}{a^2 + b^2}}.\]